设(R(alpha_{1},alpha_{2},cdots,alpha_{s})=R(beta_{1},beta_{2},cdots,beta_{t})=r)，且(alpha_{1},alpha_{2},cdots,alpha_{s}和beta_{1},beta_{2},cdots,beta_{t})均为n维向量组，则下列结论正确的是(（quad）)。
排除法，反例(alpha_{1}=(1,0,0,0),alpha_{2}=(0,1,0,0),beta_{1}=(0,0,1,0),beta_{2}=(0,0,0,1)),则排除A、B。D错误是因为向量组的等价与向量组中所含向量的个数无关。...
已知向量组(alpha_{1},alpha_{2},alpha_{3},分别可由向量组beta_{1},beta_{2},beta_{3},线性表示，即begin{equation}left{begin{aligned}alpha_{1}&=beta_{1}-beta_{2}+beta_{3}alpha_{2}&=beta_{1}+beta_{2}-beta_{3}alpha_{3}&=-beta_{1}+beta_{2}+beta_{3}end{aligned}right.end{equation}，)判别这两个向量组不等价是否正确。
这两个向量组等价。
设(n)维向量组(alpha_{1},alpha_{2},cdots,alpha_{s})的秩为(r,)则(（quad）)为错误命题。
线性无关组(alpha_{1},alpha_{2},cdots,alpha_{s})未必是(R^n)的极大无关组。